例如，在输入框中给定两个日期:
<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>
<script>
alert(datediff("day", first, second)); // what goes here?
</script>
如何在JavaScript中获得两个日期之间的天数?
当前回答
function formatDate(seconds, dictionary) {
var foo = new Date;
var unixtime_ms = foo.getTime();
var unixtime = parseInt(unixtime_ms / 1000);
var diff = unixtime - seconds;
var display_date;
if (diff <= 0) {
display_date = dictionary.now;
} else if (diff < 60) {
if (diff == 1) {
display_date = diff + ' ' + dictionary.second;
} else {
display_date = diff + ' ' + dictionary.seconds;
}
} else if (diff < 3540) {
diff = Math.round(diff / 60);
if (diff == 1) {
display_date = diff + ' ' + dictionary.minute;
} else {
display_date = diff + ' ' + dictionary.minutes;
}
} else if (diff < 82800) {
diff = Math.round(diff / 3600);
if (diff == 1) {
display_date = diff + ' ' + dictionary.hour;
} else {
display_date = diff + ' ' + dictionary.hours;
}
} else {
diff = Math.round(diff / 86400);
if (diff == 1) {
display_date = diff + ' ' + dictionary.day;
} else {
display_date = diff + ' ' + dictionary.days;
}
}
return display_date;
}
其他回答
从DatePicker小部件使用formatDate怎么样?您可以使用它来转换时间戳格式的日期(从01/01/1970开始的毫秒)，然后做一个简单的减法。
在撰写本文时，其他答案中只有一个正确处理DST(夏令时)转换。以下是位于加州的一个系统的结果:
1/1/2013- 3/10/2013- 11/3/2013-
User Formula 2/1/2013 3/11/2013 11/4/2013 Result
--------- --------------------------- -------- --------- --------- ---------
Miles (d2 - d1) / N 31 0.9583333 1.0416666 Incorrect
some Math.floor((d2 - d1) / N) 31 0 1 Incorrect
fuentesjr Math.round((d2 - d1) / N) 31 1 1 Correct
toloco Math.ceiling((d2 - d1) / N) 31 1 2 Incorrect
N = 86400000
虽然数学。round返回正确的结果，我认为它有点笨拙。相反，当DST开始或结束时，通过显式计算UTC偏移量的变化，我们可以使用精确的算术:
function treatAsUTC(date) {
var result = new Date(date);
result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
return result;
}
function daysBetween(startDate, endDate) {
var millisecondsPerDay = 24 * 60 * 60 * 1000;
return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}
alert(daysBetween($('#first').val(), $('#second').val()));
解释
JavaScript的日期计算很棘手，因为date对象内部存储的时间是UTC，而不是本地时间。例如，3/10/2013太平洋标准时间12:00 AM (UTC-08:00)存储为3/10/2013上午8:00 UTC, 3/11/2013太平洋夏令时12:00 AM (UTC-07:00)存储为3/11/2013上午7:00 UTC。在这一天，从午夜到午夜，当地时间在UTC只有23小时!
虽然本地时间中的一天可以大于或小于24小时，但国际标准时间中的一天总是24小时上面所示的daysBetween方法利用了这一事实，它首先调用treatAsUTC将本地时间调整为午夜UTC，然后再进行减法和除法。
1. JavaScript忽略闰秒。
function timeDifference(date1, date2) { var oneDay = 24 * 60 * 60; // hours*minutes*seconds var oneHour = 60 * 60; // minutes*seconds var oneMinute = 60; // 60 seconds var firstDate = date1.getTime(); // convert to milliseconds var secondDate = date2.getTime(); // convert to milliseconds var seconds = Math.round(Math.abs(firstDate - secondDate) / 1000); //calculate the diffrence in seconds // the difference object var difference = { "days": 0, "hours": 0, "minutes": 0, "seconds": 0, } //calculate all the days and substract it from the total while (seconds >= oneDay) { difference.days++; seconds -= oneDay; } //calculate all the remaining hours then substract it from the total while (seconds >= oneHour) { difference.hours++; seconds -= oneHour; } //calculate all the remaining minutes then substract it from the total while (seconds >= oneMinute) { difference.minutes++; seconds -= oneMinute; } //the remaining seconds : difference.seconds = seconds; //return the difference object return difference; } console.log(timeDifference(new Date(2017,0,1,0,0,0),new Date()));
使用以下公式可以计算出两个不同TZs的休止日期之间的完整证明天数差:
var start = new Date('10/3/2015'); var end = new Date('11/2/2015'); Var天数= (end - start) / 1000 / 60 / 60 / 24; console.log(天); //实际上是30;但由于夏令时，将显示31.0xxx //你需要抵消如下 days = days - (end.getTimezoneOffset() - start.getTimezoneOffset()) / (60 * 24); console.log(天);
我来找这个小工具在里面你会找到这个的函数。这里有一个简短的例子:
<script type="text/javascript" src="date.js"></script>
<script type="text/javascript">
var minutes = 1000*60;
var hours = minutes*60;
var days = hours*24;
var foo_date1 = getDateFromFormat("02/10/2009", "M/d/y");
var foo_date2 = getDateFromFormat("02/12/2009", "M/d/y");
var diff_date = Math.round((foo_date2 - foo_date1)/days);
alert("Diff date is: " + diff_date );
</script>
